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Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

(I) Select the correct option from the given alternatives.

Question 1.
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 2
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) if A – λI is a singular matrix, then ___________
(a) λ = 0
(b) λ² – 3λ – 4 = 0
(c) λ² + 3λ – 4 = 0
(d) λ² – 3λ – 6 = 0
Answer:
(b) λ² – 3λ – 4 = 0
Hint:

Question 2.
Consider the matrices A = \(\left[\begin{array}{ccc}
4 & 6 & -1 \\
3 & 0 & 2 \\
1 & -2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
0 & 1 \\
-1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\). Out of the given matrix products, ___________
(i) (AB)^TC
(ii) C^TC(AB)^T
(iii) C^TAB
(iv) A^TABB^TC
(a) Exactly one is defined
(b) Exactly two are defined
(c) Exactly three are defined
(d) all four are defined
Answer:
(c) Exactly three are defined
Hint:
A is of order 3 × 3, B is of order 3 × 2 and C is of order 3 × 1.
(AB)^TC is of order 2 × 1.
C^TC and (AB)^T are of different orders.
C^TC (AB)^T is not defined.
C^TAB is of order 1 × 2.
A^TABB^TC is of order 3 × 1.

Question 3.
If A and B are square matrices of equal order, then which one is correct among the following?
(a) A + B = B + A
(b) A + B = A – B
(c) A – B = B – A
(d) AB = BA
Answer:
(a) A + B = B + A
Hint:
Matrix addition is commutative.
∴ A + B = B + A

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]\) is a matrix satisfying the equation AA^T = 9I, where I is the identity matrix of order 3, then the ordered pair (a, b) is equal to ___________
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Answer:
(d) (-2, -1)

Question 5.
If A = \(\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right]\) and |A³| = 125, then α = ___________
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
(a) ±3
Hint:
|A³| = 125
|A|³ = 53 …….[∵ |Aⁿ| = |A|ⁿ, n ∈ N]
∴ |A| = 5
\(\left|\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right|=5\)
α² – 4 = 5
α² = 9
∴ α = ± 3

Question 6.
If \(\left[\begin{array}{ll}
5 & 7 \\
x & 1 \\
2 & 6
\end{array}\right]-\left[\begin{array}{cc}
1 & 2 \\
-3 & 5 \\
2 & y
\end{array}\right]=\left[\begin{array}{cc}
4 & 5 \\
4 & -4 \\
0 & 4
\end{array}\right]\), then ___________
(a) x = 1, y = -2
(b) x = -1, y = 2
(c) x = 1, y = 2
(d) x = -1, y = -2
Answer:
(c) x = 1, y = 2

Question 7.
If A + B = \(\left[\begin{array}{ll}
7 & 4 \\
8 & 9
\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\), then the value of A is ___________
(a) \(\left[\begin{array}{ll}
3 & 1 \\
4 & 3
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
6 & 2 \\
8 & 6
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
7 & 6 \\
8 & 12
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)

Question 8.
If \(\left[\begin{array}{cc}
x & 3 x-y \\
z x+z & 3 y-w
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
4 & 7
\end{array}\right]\), then ___________
(a) x = 3, y = 7, z = 1, w = 14
(b) x = 3, y = -5, z = -1, w = -4
(c) x = 3, y = 6, z = 2, w = 7
(d) x = -3, y = -7, z = -1, w = -14
Answer:
(a) x = 3, y = 7, z = 1, w = 14

Question 9.
For suitable matrices A, B, the false statement is ___________
(a) (AB)^T = A^TB^T
(B) (A^T)^T = A
(C) (A – B)^T = A^T – B^T
(D) (A + B)^T = A^T + B^T
Answer:
(a) (AB)^T = A^TB^T
Hint:
(AB)^T = B^TA^T

Question 10.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
0 & 3
\end{array}\right]\) and f(x) = 2x² – 3x, then f(A) = ___________
(a) \(\left[\begin{array}{cc}
14 & 1 \\
0 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
-14 & 1 \\
0 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-14 & -1 \\
0 & -9
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
Hint:

(II) Answer the following questions.

Question 1.
If A = diag[2, -3, -5], B = diag[4, -6, -3] and C = diag [-3, 4, 1], then find
i. B + C – A
ii. 2A + B – 5C.
Solution:

Question 2.
If f(α) = A = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), find
i. f(-α)
ii. f(-α) + f(α)
Solution:

Question 3.
Find matrices A and B, where
(i) 2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) and A + 3B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\)
(ii) 3A – B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 0 & 5
\end{array}\right]\) and A + 5B = \(\left[\begin{array}{ccc}
0 & 0 & 1 \\
-1 & 0 & 0
\end{array}\right]\)
Solution:
i. Given equations are
2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) …….(i)

Question 4.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right]\), verify
i. (A + 2B^T)^T = A^T + 2B
ii. (3A – 5B^T)^T = 3A^T – 5B.
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A^T = I, where I is a unit matrix of order 2 × 2, then find the value of α.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & 2 \\
-1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & -1 & -3
\end{array}\right]\), show that AB is singular.
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\), show that AB and BA are both singular matrices.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\), show that BA = 6I.
Solution:

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -2
\end{array}\right]\), verify that |AB| = |A|.|B|.
Solution:

Question 10.
If Aα = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that Aα . Aβ = Aα+β
Solution:

Question 11.
If A = \(\left[\begin{array}{cc}
1 & \omega \\
\omega^{2} & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\omega^{2} & 1 \\
1 & \omega
\end{array}\right]\), where ω is a complex cube root of unity, then show that AB + BA + A – 2B is a null matrix.
Solution:
ω is the complex cube root of unity.
ω³ = 1
ω³ – 1 = 0
(ω – 1) (ω² + ω + 1) = 0
ω = 1 or ω² + ω + 1 = 0
But, ω is a complex number.
1 + ω + ω² = 0 …….(i)
AB + BA + A – 2B

which is a null matrix.

Question 12.
If A = \(\left[\begin{array}{lrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\), show that A² = A.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right]\), show that A² = I.
Solution:

Question 14.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A² – 5A – 14I = 0.
Solution:

Question 15.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), show that A – 4A + 3I = 0.
Solution:

Question 16.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & -4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & x \\
y & 0
\end{array}\right]\) and (A + B)(A – B) = A² – B², find x and y.
Solution:
(A + B)(A – B) = A² – B²
A² – AB + BA – B² = A² – B²
-AB + BA = 0
AB = BA

By equality of matrices, we get
2 – 4x = -3x
∴ x = 2 and 2y = 2x
y = x
∴ y = 2
∴ x = 2, y = 2

Question 17.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\), show that (A + B)(A – B) ≠ A² – B².
Solution:
We have to prove that
(A + B) . (A – B) ≠ A² – B²
i.e., to prove that A(A – B) + B(A – B) ≠ A² – B²
i.e., to prove that A² – AB + BA – B² ≠ A² – B²
i.e., to prove that AB ≠ BA.

∴ AB ≠ BA

Question 18.
If A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & -2
\end{array}\right]\), find A³.
Solution:

∴ A² = I
Multiplying throughout by A, we get
A³ = A . I
∴ A³ = A

Question 19.
Find x, y if,

Solution:

Question 20.
Find x, y, z if

Solution:

Question 21.
If A = \(\left[\begin{array}{ccc}
2 & 1 & -3 \\
0 & 2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
3 & -1 & 4
\end{array}\right]\), find AB^T and A^TB.
Solution:

Question 22.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\), show that (AB)^T = B^TA^T.
Solution:

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for all n ∈ N.
Solution:

Question 24.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), for all n ∈ N.
Solution:

Question 25.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rupees) of these crops by both the farmers for the month of April and may 2008 is given below,

Find
(i) the total sale in rupees for two months of each farmer for each crop.
(ii) the increase in sales from April to May for every crop of each farmer.
Solution:
(i) Total sale for Shantaram:
For rice = 15000 + 18000 = ₹ 33000.
For wheat = 13000 + 15000 = ₹ 28000.
For groundnut = 12000 + 12000 = ₹ 24000.
Total sale for Kantaram:
For rice = 18000 + 21000 = ₹ 39000
For wheat = 15000 + 16500 = ₹ 31500
For groundnut = 8000 + 16000 = ₹ 24000

Alternate method:
Matrix form

∴ The total sale of April and May of Shantaram in ₹ is ₹ 33000 (rice), ₹ 28000 (wheat), ₹ 24000 (groundnut), and that of Kantaram in ₹ is ₹ 39000(rice), ₹ 31500(wheat), and ₹ 24000 (groundnut).

(ii) Increase in sale from April to May for Shantaram:
For rice = 18000 – 15000 = ₹ 3000
For wheat = 15000 – 13000 = ₹ 2000
For groundnut = 12000 – 12000 = ₹ 0
Increase in sale from April to May for Kantaram:
For rice = 21000 – 18000 = ₹ 3000
For wheat = 16500 – 15000 = ₹ 1500
For groundnut = 16000 – 8000 = ₹ 8000

Alternate method:
Matrix form

∴ The increase in sales for Shantaram from April to May in each crop is ₹ 3000 (rice), ₹ 2000(wheat), 0 (groundnut), and that for Kantaram is ₹ 3000 (rice), ₹ 1500 (wheat), and ₹ 8000 (groundnut).

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7

Question 1.
Find A^T, if
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
2 & -6 & 1 \\
-4 & 0 & 5
\end{array}\right]\)
Solution:
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
1 & -4 \\
3 & 5
\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}
-4 & 0 & 5
2 & -6 & 1 \\
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\)

[Note: Answer given in the textbook is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
6 & 0 \\
1 & 5
\end{array}\right]\). However, as per our calculation it is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\). ]

Question 2.
If [a_ij]_3×3 where a_ij = 2(i – j), find A and
A^T. State whether A and A^T are symmetric or skew-symmetric matrices?
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given a_ij = 2 (i — j)
∴ a₁₁ = 2(1-1) = 0,
a₁₂ = 2(1-2) = -2,
a₁₃ = 2(1-3) = -4,
a₂₁ = 2(2-1) = 2,
a₂₂ = 2(2-2) = 0,
a₂₃=2(2-3) = -2,
a₃₁ = 2(3-1) = 4,
a₃₂ = 2(3-2) = 2,
a₃₃=2(3-3) = 0

∴ A^T = -A and A = -A^T
∴ A and A^T both are skew-symmetric matrices.

Questionn 3.
If A = \(\left[\begin{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right]\), prove that (2A)^T = 2A^T.
Solution:

From (i) and (ii), we get
(2A)^T = 2A^T

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\), prove that (3A)^T = 3A^T.
Solution:

From (i) and (ii), we get
(3A)^T = 3A^T

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 1+2 i & 1-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\),
where i = \(\sqrt{-1}\), prove that A^T = – A.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\) , B = \(\left[\begin{array}{cc}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 4 \\
-2 & 3
\end{array}\right]\) then show that
i. (A + B)^T = AT + B^T
ii. (A – C)^T = A^T – C^T
Solution:

From (i) and (ii), we get
(A + B)^T = AT + B^T
[Note: The question has been modified.]

From (i) and (ii), we get
(A – C)^T = A^T – C^T</sup

Question 7.
If A = \(\left[\begin{array}{cc}
5 & 4 \\
-2 & 3
\end{array}\right]\) and \(\left[\begin{array}{cc}
-1 & 3 \\
4 & -1
\end{array}\right]\) then find C^T, such that 3A – 2B + C = I, where I is the unit matrix of order 2.
Solution:
3A – 2B + C = I
∴ C = I + 2B – 3A

Question 8.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
i. A^T + 4B^T
ii. 5A^T – 5B^T
Solution:

ii. ii. 5A^T – 5B^T = 5(A^T – B^T)

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
2 & 1 & -4 \\
3 & 5 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
0 & 2 & 3 \\
-1 & -1 & 0
\end{array}\right]\), verify that (A + 2B + 3C)^T = A^T + 2B^T + 3C^T.
Solution:
A + 2B + 3C

∴ A^T + 2B^T + 3C^T = \(\left[\begin{array}{cc}
5 & 6 \\
8 & 8 \\
2 & -2
\end{array}\right]\)
From (i) and (ii), we get
(A + 2B + 3C)^T = A^T + 2B^T + 3C^T

Question 10.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right]\), prove that (A + B^T)^T = A^T + B.
prove that (A + B^T)^T = A^T + B
Solution:


From (i) and (ii), we get
(A + B^T)^T = A^T + B

Question 11.
Prove that A + A^T is a symmetric and A – A^T is a skew symmetric matrix, where
i. A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 2 & 1 \\
-2 & -3 & 2
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
5 & 2 & -4 \\
3 & -7 & 2 \\
4 & -5 & -3
\end{array}\right]\)
Solution:

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.


∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.

∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

Question 12.
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
i. \(\left[\begin{array}{cc}
4 & -2 \\
3 & -5
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as
A = \(\frac{1}{2}\) (A + A^T) + \(\frac{1}{2}\) (A – A^T)

P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
A = P + Q
A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

∴ P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
∴ A = P + Q
∴ A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

Question 13.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\), verify that
i. (AB)^T = B^TA^T
ii. (BA)^T = A^TB^T
Solution:


From (i) and (ii), we get
(AB)^T = B^TA^T

From (i) and (ii) we get
(BA)^T = A^TB^T

Question 14.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A^TA = I, where I is the unit matrix of order 2.
Solution:


∴ A^TA = I, where I is the unit matrix of order 2.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.5

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\)
Show that
i. A+B=B+A
ii. (A + B) + C = A + (B + C)
Solution:

From (i) and (ii), we get
A + B = B + A

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\) then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:

Question 3.
If A = \(\), B = \(\) then find the matrix C such that A + B + C is a zero matrix.
Solution:
A+ B + C is a zero matrix.
∴ A + B + C = O
C = -(A + B)

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\) B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\) , find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A-4B + 5X = C
∴ 5X = C + 4B – 3A

Question 5.
Solve the following equations for X and Y, if 3X – Y = \(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:
Given equations are
\(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\)……………….. (i)
and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\) ………………(ii)
By (i) x 3 – (ii) we get

Question 6.
Find the matrices A and B, if 2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:
Given equations are
2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) ……………….. (i)
and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\) ……………….(ii)
By (i) – (ii) x 2, we get

Question 7.
Simplify \(\cos \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:

Quesiton 8.
If A = \(\left[\begin{array}{cc}
1 & 2 i \\
-3 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 i & 1 \\
2 & -3
\end{array}\right]\) where i =\(\sqrt{-1}\), find A + B and A – B. Show that A + B is singular. Is A – B singular? Justify your answer.
Solution:

Question 9.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
2x + y – 1 = 3 and 4y = 18
∴ 2x + y = 4 and y = \(\frac{18}{4}=\frac{9}{2}\)
∴ 2x + \(\frac{9}{2}\) = 4
∴ 2x = 4 – \(\frac{9}{2}\)
∴ 2x = \(\frac{1}{2}=\)
∴ x = –\(\frac{1}{4}=\) and y = \(\frac{9}{2}=\)

Question 10.
If \(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)

∴ By equality of matrices, we get
2a + b = 2 ….(i)
3a – b = 3 ….(ii)
c + 2d = 4 ….(iii)
2c – d = -1 ….(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (iii), we get
\(\frac{2}{5}\) + 2d = 4
∴ 2d = 4 – \(\frac{2}{5}\)
∴ 2d = \(\frac{18}{5}\)
∴ d = \(\frac{9}{5}\)
[Note: Answer given in the textbook is d = \(\frac{3}{5}\).
However, as per our calculation it is d = \(\frac{9}{5}\).]

Question 11.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B.

i. Find the increase in sales in Rupees from July to August 2017.
ii. If both book shops got 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.
Solution:
i. Increase in sales in rupees from July to August 2017
For Suresh:
Increase in sales for Physics books
= 6650 – 5600= ₹ 1050
Increase in sales for Chemistry books
= 7055 – 6750 = ₹ 305
Increase in sales for Mathematics books
= 8905 – 8500 = ₹ 405
For Ganesh:
Increase in sales for Physics books
= 7000 – 6650 = ₹ 350
Increase in sales for Chemistry books
= 7500 – 7055 = ₹ 445
Increase in sales for Mathematics books
= 10200 – 8905 = ₹ 1295
[Note: Answers given in the textbook are 1760, 2090. However, as per our calculation they are 1050, 305, 405, 350, 445, 1295.]

ii. Both book shops got 10% profit in the month of August 2017.
For Suresh:
Profit for Physics books = \(\frac{6650 \times 10}{100}\) = ₹ 665
Profit for Chemistry books = \(\frac{7055 \times 10}{100}\) = ₹ 705.50
Profit for Mathematics books = \(\frac{8905 \times 10}{100}\) = ₹ 890.50

For Ganesh:
Profit for Physics books = \(\frac{7000 \times 10}{100}\) = ₹ 700
Profit for Chemistry books = \(\frac{7500 \times 10}{100}\) = ₹ 750
Profit for Mathematics books = \(\frac{10200 \times 10}{100}\) = ₹ 1020
[Note: Answers given in the textbook for Suresh’s profit in Chemistry and Mathematics books are ? 675 and ?850 respectively. However, as per our calculation profit amounts are ₹ 705.50 and ₹ 890.50 respectively.]

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.4

Question 1.
Construct a matrix A = [a_ij]_{3 x 2} whose elements ay are given by
i. a_ij = \(\frac{(\mathbf{i}-\mathbf{j})^{2}}{5-\mathbf{i}}\)
ii. a_ij = i – 3j
iii. a_ij \(\frac{(i+j)^{3}}{5}\)
Solution:

[Note: Answer given in the textbook is A = \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{2} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{3} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\) ].

ii. a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = -2,
a₁₂= 1 – 3(2) = 1 – 6 = -5,
a₂₁ = 2 – 3(1) = 2 – 3 =-1,
a₂₂ = 2 – 3(2) = 2 – 6 = – 4
a₃₁ = 3 – 3(1) = 3-3 = 0,
a₃₂ = 3 – 3(2) = 3 – 6 = -3
∴ A = \(\left[\begin{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
\end{array}\right]\)

iii. a_ij = \(\frac{(i+j)^{3}}{5}\)

Question 2.
Classify the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular, a symmetric or a skew- symmetric matrix.
i. \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
As every element below the diagonal is zero in matrix A.
∴ A is an upper triangular matrix.

ii. \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{ccc}
0 & -4 & -7 \\
4 & 0 & 3 \\
7 & -3 & 0
\end{array}\right]\)
∴ A^T = \(-\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

iii. \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
∴ As matrix A has only one column.
∴ A is a column matrix.

iv. \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
As matrix A has only one row.
∴ A is a row matrix.

v. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero and diagonal elements same.
∴ A is a scalar matrix.

vi. \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
As every element above the diagonal is zero in matrix A.
∴ A is a lower triangular matrix.

vii. \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero.
∴ A is a diagonal matrix.

viii. \(\left[\begin{array}{ccc}
10 & -15 & 27 \\
-15 & 0 & \sqrt{34} \\
27 & \sqrt{34} & \frac{5}{3}
\end{array}\right]\)
Solution:

∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.

ix. \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
In matrix A, all the non-diagonal elements are zero and diagonal elements are one.
∴ A is a unit (identity) matrix.

x. \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.
∴ A is a symmetric matrix.

Question 3.
Which of the following matrices are singular or non-singular?
i. \(\left[\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{p} & \mathbf{q} & \mathbf{r} \\
\mathbf{2 a}-\mathbf{p} & \mathbf{2 b}-\mathbf{q} & \mathbf{2 c}-\mathbf{r}
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
iv. \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
Solution:

ii. Let A = \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right|\)
Applying C₂ → C₂ + C₁
|A| = \(\left|\begin{array}{ccc}
5 & 5 & 5 \\
1 & 100 & 100 \\
6 & 105 & 105
\end{array}\right|\)
= 0 … [∵ C₂ and C₃ are identical]
∴ A is a singular matrix.

iii. Let A = \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right| \)
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49 = 52 ≠ 0
∴ A is a non-singular matrix.

iv. Let A = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\) = 49 + 20 = 69 ≠ 0

Question 4.
Find k, if the following matrices are singular.
i. \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
Since A is a singular matrix,
|A|=0
∴ \(\left|\begin{array}{cc}
7 & 3 \\
-2 & \mathrm{k}
\end{array}\right|\) = o
∴ 7k + 6 = 0
∴ 7k = -6
k = -6/7

ii. Let A = \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
Since A is a singular matrix,
|A|= 0
∴ \(\left|\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{k} & 1 \\
10 & 9 & 1
\end{array}\right|\) = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6

iii. Let A = \(\left[\begin{array}{ccc}
\mathbf{k}-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Since A is a singular matrix
|A| = 0
∴ \(\left|\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right|\)
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3 (-6 – 1) = 0
∴ 8k-8-20-21 =0
∴ 8k = 49
∴ k = 49/8

Question 5.
If A = \(\left[\begin{array}{lll}
5 & 1 & -1 \\
3 & 2 & 0
\end{array}\right]\), find (A^T)^T.
Solution:

Question 6.
If A = \(\), find (A^T)^T.
Solution:

Question 7.
Find a, b, c, if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symmetric matrix.
Solution:

Question 8.
Find x, y, z, if \(\) is a symmetric matrix.
Solution:

Question 9.
For each of the following matrices, using its transpose, state whether it is symmetric, skew-symmetric or neither.
i. \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
0 & 1+2 \mathbf{i} & \mathbf{i}-2 \\
-1-2 \mathbf{i} & 0 & -7 \\
2-\mathbf{i} & 7 & 0
\end{array}\right]\)
Solution:
i. Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T =\(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T = A, i.e., A = A^T
∴ A is a symmetric matrix.

ii.

∴ A ≠ A^T, i.e., A ≠ -A^T
∴ A is neither a symmetric nor skew-symmetric matrix.

iii.

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [a_ij]_{3 x 3}, where a_ij = i – j. State whether A is symmetric or skew-symmetric.
Solution:
A = [a_ij]_{3 x 3}
∴ A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given, a_ij = i – j
a₁₁ = 1-1 = 0, a₁₂ = 1-2 = – 1, a₁₃ = 1 – 3 = – 2,
a₂₁ – 2 – 1 = 1, a₂₂ = 2 – 2 = 0, a₂₃ =2 – 3 = – 1,
a₃₁ = 3 – 1 = 2, a₃₂ = 3 – 2 = 1, a₃₃ = 3 – 3 = 0

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A)

I. Select the correct option from the given alternatives.

Question 1.
The determinant D = \(\left|\begin{array}{ccc}
a & b & a+b \\
b & c & b+c \\
a+b & b+c & 0
\end{array}\right|\) = 0, if
(a) a, b, c are in A.P.
(b) a, b, c are in G.P.
(c) a, b, c are in H.P.
(d) α is a root of ax² + 2bx + c = 0
Answer:
(b) a, b, c are in G.P.
Hint:
Applying R₃ → R₃ – (R₁ + R₂), we get
\(\left|\begin{array}{llc}
a & b & a+b \\
b & c & b+c \\
0 & 0 & -(a+2 b+c)
\end{array}\right|=0\)
∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0
∴ (-ac + b²) (a + 2b + c) = 0
∴ -ac + b² = 0 or a + 2b + c = 0
∴ b² = ac
∴ a, b, c are in G.P.

Question 2.
If \(\left|\begin{array}{lll}
x^{k} & x^{k+2} & x^{k+3} \\
y^{k} & y^{k+2} & y^{k+3} \\
z^{k} & z^{k+2} & z^{k+3}
\end{array}\right|\) = (x – y) (y – z) (z – x) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) then
(a) k = -3
(b) k = -1
(c) k = 1
(d) k = 3
Answer:
(b) k = -1
Hint:

Question 3.
Let D = \(\left|\begin{array}{ccc}
\sin \theta \cdot \cos \phi & \sin \theta \cdot \sin \phi & \cos \theta \\
\cos \theta \cdot \cos \phi & \cos \theta \cdot \sin \phi & -\sin \theta \\
-\sin \theta \cdot \sin \phi & \sin \theta \cdot \cos \phi & 0
\end{array}\right|\) then
(a) D is independent of θ
(b) D is independent of φ
(c) D is a constant
(d) \(\frac{d D}{d}\) at θ = \(\frac{\pi}{2}\) is equal to 0
Answer:
(b) D is independent of φ

Question 4.
The value of a for which the system of equations a³x + (a + 1)y + (a + 2)³ z = 0, ax + (a + 1)y + (a + 2)z = 0 and x + y + z = 0 has a non zero solution is
(a) 0
(b) -1
(c) 1
(d) 2
Answer:
(b) -1
Hint:
The given system of equations will have a non-zero solution, if

Question 5.
\(\left|\begin{array}{lll}
b+c & c+a & a+b \\
q+r & r+p & p+q \\
y+z & z+x & x+y
\end{array}\right|=\)
(a) 2 \(\left|\begin{array}{lll}
c & b & a \\
r & q & p \\
z & y & x
\end{array}\right|\)
(b) 2 \(\left|\begin{array}{lll}
b & a & c \\
q & p & r \\
y & x & z
\end{array}\right|\)
(c) 2 \(\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
(d) 2 \(\left|\begin{array}{lll}
a & c & b \\
p & r & q \\
x & z & y
\end{array}\right|\)
Answer:
(c) 2 \(\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
Hint:

Question 6.
The system 3x – y + 4z = 3, x + 2y – 3z = -2 and 6x + 5y + λz = -3 has atleast one solution when
(a) λ = -5
(b) λ = 5
(c) λ = 3
(d) λ = -13
Answer:
(a) λ = -5
Hint:
The given system of equations will have more than one solution if

Question 7.
If x = -9 is a root of \(\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|=0\), has other two roots are
(a) 2, -7
(b) -2, 7
(c) 2, 7
(d) -2, -7
Answer:
(c) 2, 7
Hint:

Question 8.
If \(\left|\begin{array}{ccc}
6 i & -3 i & 1 \\
4 & 3 i & -1 \\
20 & 3 & i
\end{array}\right|\) = x + iy, then
(a) x = 3, y = 1
(b) x = 1, y = 3
(c) x = 0, y = 3
(d) x = 0, y = 0
Answer:
(d) x = 0, y = 0

Question 9.
If A(0, 0), B(1, 3) and C(k, 0) are vertices of triangle ABC whose area is 3 sq.units, then the value of k is
(a) 2
(b) -3
(c) 3 or -3
(d) -2 or 2
Answer:
(d) -2 or 2

Question 10.
Which of the following is correct?
(a) Determinant is a square matrix
(b) Determinant is number associated to matrix
(c) Determinant is a number associated with a square matrix
(d) None of these
Answer:
(c) Determinant is a number associated with a square matrix

II. Answer the following questions.

Question 1.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)
Solution:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)

= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)
= 2(4) + 5(1) + 7(-18)
= 8 + 5 – 126
= -113

(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

Question 2.
Evaluate determinant along second column \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 2 & -2 \\
0 & 1 & -2
\end{array}\right|\)
Solution:

Question 3.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & 3 & 5 \\
400 & 600 & 1000 \\
48 & 47 & 18
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
101 & 102 & 103 \\
106 & 107 & 108 \\
1 & 2 & 3
\end{array}\right|\)
by using properties.
Solution:

Question 4.
Find the minors and cofactors of elements of the determinants.
(i) \(\left|\begin{array}{ccc}
-1 & 0 & 4 \\
-2 & 1 & 3 \\
0 & -4 & 2
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right|\)
Solution:

Question 5.
Find the values of x, if
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
⇒ 1(-10x² + 10x) – 4(5x² + 5) + 20(2x + 2) = 0
⇒ -10x² + 10x – 20x² – 20 + 40x + 40 = 0
⇒ -30x² + 50x + 20 = 0
⇒ 3x² – 5x – 2 = 0 …..[Dividing throughout by (-10)]
⇒ 3x² – 6x + x – 2 = 0
⇒ 3x(x – 2) + 1(x – 2) = 0
⇒ (x – 2) (3x + 1) = 0
⇒ x – 2 = 0 or 3x + 1 = 0
⇒ x = 2 or x = \(-\frac{1}{3}\)

(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
⇒ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
⇒ 1(-12) – 2x(-15) + 4x(-3) = 0
⇒ -12 + 30x – 12x = 0
⇒ 18x = 12
⇒ x = \(\frac{12}{18}=\frac{2}{3}\)

Question 6.
By using properties of determinant, prove that \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:

Question 7.
Without expanding the determinants, show that

Solution:

Question 8.
If \(\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0\) then show that \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1\)
Solution:

Question 9.
Solve the following linear equations by Cramer’s Rule.
(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1
(ii) \(\frac{1}{x}+\frac{1}{y}=\frac{3}{2}, \quad \frac{1}{y}+\frac{1}{z}=\frac{5}{6}, \quad \frac{1}{z}+\frac{1}{x}=\frac{4}{3}\)
(iii) 2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3
(iv) x + y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12
Solution:
(i) Given equations are
2x – y + z = 1
x + 2y + 3z = 8
3x + y – 4z = 1
D = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 – 13 – 5
= -40 ≠ 0

Dx = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
\end{array}\right|\)
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40

Dy = \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-32 – 3) -1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80

Dz = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
\end{array}\right|\)
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,

∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

(ii) Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
p + q = \(\frac{3}{2}\)
i.e., 2p + 2q = 3
i.e., 2p + 2q + 0 = 3
q + r = \(\frac{5}{6}\)
i.e., 6q + 6r = 5,
i.e., 0p + 6q + 6r = 5
r + p = \(\frac{4}{3}\)
i.e., 3r + 3p = 4,
i.e., 3p + 0q + 3r = 4
D = \(\left|\begin{array}{lll}
2 & 2 & 0 \\
0 & 6 & 6 \\
3 & 0 & 3
\end{array}\right|\)
= 2(18 – 0) -2(0 – 18) + 0
= 2(18) – 2(-18)
= 36 + 36
= 72 ≠ 0

Dp = \(\left|\begin{array}{lll}
3 & 2 & 0 \\
5 & 6 & 6 \\
4 & 0 & 3
\end{array}\right|\)
= 3(18 – 0) – 2(15 – 24) + 0
= 3(18) – 2(-9)
= 54 + 18
= 72

Dq = \(\left|\begin{array}{lll}
2 & 3 & 0 \\
0 & 5 & 6 \\
3 & 4 & 3
\end{array}\right|\)
= 2(15 – 24) – 3(0 – 18) + 0
= 2(-9) – 3(-18)
= -18 + 54
= 36

Dr = \(\left|\begin{array}{lll}
2 & 2 & 3 \\
0 & 6 & 5 \\
3 & 0 & 4
\end{array}\right|\)
= 2(24 – 0) – 2(0 – 15) + 3(0 – 18)
= 2(24) – 2(-15) + 3(-18)
= 48 + 30 – 54
= 24
By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

(iii) Given equations are
2x + 3y + 3z = 5
x – 2y + z = -4
3x – y – 2z = 3
D = \(\left|\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\)
= 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6)
= 2(5) – 3(-5) + 3(5)
= 10 + 15 + 15
= 40 ≠ 0

Dx = \(\left|\begin{array}{ccc}
5 & 3 & 3 \\
-4 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\)
= 5(4 + 1) – 3(8 – 3) + 3(4 + 6)
= 5(5) – 3(5) + 3(10)
= 25 – 15 + 30
= 40

Dy = \(\left|\begin{array}{ccc}
2 & 5 & 3 \\
1 & -4 & 1 \\
3 & 3 & -2
\end{array}\right|\)
= 2(8 – 3) – 5(-2 – 3) + 3(3 + 12)
= 2(5) – 5(-5) + 3(15)
= 10 + 25 + 45
= 80

Dz = \(\left|\begin{array}{ccc}
2 & 3 & 5 \\
1 & -2 & -4 \\
3 & -1 & 3
\end{array}\right|\)
= 2(-6 – 4) – 3(3 + 12) + 5(-1 + 6)
= 2(-10) – 3(15) + 5(5)
= -20 -45 + 25
= -40
By Cramer’s Rule,

∴ x = 1, y = 2 and z = -1 are the solutions of the given equations.

(iv) Given equations are
x – y + 2z = 7
3x + 4y – 5z = 5
2x – y + 3z = 12
D = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right|\)
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0

Dx = \(\left|\begin{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
\end{array}\right|\)
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7) + 1(75) + 2(-53)
= 49 + 75 – 106
= 18

Dy = \(\left|\begin{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
\end{array}\right|\)
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6

Dz = \(\left|\begin{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
\end{array}\right|\)
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53) + 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,

∴ x = \(\frac{9}{2}\), y = \(\frac{-3}{2}\) and z = \(\frac{1}{2}\) are the solutions of the given equations.

Question 10.
Find the value of k, if the following equations are consistent.
(i) (k + 1)x + (k – 1)y + (k – 1) = 0
(k – 1)x + (k + 1)y + (k – 1) = 0
(k – 1)x + (k – 1)y + (k + 1) = 0
(ii) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3
(iii) (k – 2)x + (k – 1)y = 17, (k – 1)x +(k – 2)y = 18 and x + y = 5
Solution:
(i) Given equations are
(k + 1)x + (k – 1)y + (k – 1) = 0
(k – 1)x + (k + 1)y + (k – 1) = 0
(k – 1)x + (k – 1)y + (k + 1) = 0
Since these equations are consistent,

⇒ 2(2k + 2 + 2k – 2) – 0 + (k – 1) (4 – 0) = 0
⇒ 2(4k) + (k – 1)4 = 0
⇒ 8k + 4k – 4 = 0
⇒ 12k – 4 = 0
⇒ k = \(\frac{4}{12}=\frac{1}{3}\)

(ii) Given equations are
3x + y – 2 = 0
kx + 2y – 3 = 0
2x – y = 3, i.e., 2x – y – 3 = 0.
Since these equations are consistent,
\(\left|\begin{array}{rrr}
3 & 1 & -2 \\
k & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0\)
⇒ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0
⇒ 3(-9) – 1(-3k + 6) – 2(-k – 4) = 0
⇒ -27 + 3k – 6 + 2k + 8 = 0
⇒ 5k – 25 = 0
⇒ k = 5

(iii) Given equations are
(k – 2)x + (k – 1)y = 17
⇒ (k – 2)x + (k – 1)y – 17 = 0
(k – 1)x + (k – 2)y = 18
⇒ (k – 1)x + (k – 2)y – 18 = 0
x + y = 5
⇒ x + y – 5 = 0
Since these equations are consistent,

⇒ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0
⇒ -1(-5k + 28) – 1(-5k + 23) + 1(1) = 0
⇒ 5k – 28 + 5k – 23 + 1 = 0
⇒ 10k – 50 = 0
⇒ k = 5

Question 11.
Find the area of triangle whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
(ii) P(3, 6), Q(-1, 3), R(2, -1)
(iii) L(1, 1), M(-2, 2), N(5, 4)
Solution:
(i) Here, A(x₁, y₁) = A(-1, 2)
B(x₂, y₂) = B(2, 4)
C(x₃, y₃) = C(0, 0)

Since area cannot be negative,
A(ΔABC) = 4 sq.units

(ii) Here, P(x₁, y₁) = P(3, 6)
Q(x₂, y₂) = Q(-1, 3)
R(x₃, y₃) = R(2, -1)

A(ΔPQR) = \(\frac{25}{2}\) sq.units

(iii) Here, L(x₁, y₁) = L(1, 1)
M(x₂, y₂) = M(-2, 2)
N(x₃, y₃) = N(5, 4)

Since area cannot be negative,
A(ΔLMN) = \(\frac{13}{2}\) sq.units

Question 12.
Find the value of k,
(i) if the area of a triangle is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
(ii) if area of triangle is \(\frac{33}{2}\) square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Solution:
(i) Here, P(x₁, y₁) = P(k, 0)
Q(x₂, y₂) = Q(4, 0)
R(x₃, y₃) = R(0, 2)
A(ΔPQR) = 4 sq.units

(ii) Here, L(x₁, y₁) = L(3, -5), M(x₂, y₂) = M(-2, k), N(x₃, y₃) = N(1, 4)
A(ΔLMN) = \(\frac{33}{2}\) sq. units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
\(\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
-2 & k & 1 \\
1 & 4 & 1
\end{array}\right|\)
⇒ \(\pm \frac{33}{2}=\frac{1}{2}\) [3(k – 4) – (-5) (-2 – 1) + 1 (-8 – k)]
⇒ ±33 = 3k – 12 – 15 – 8 – k
⇒ ±33 = 2k – 35
⇒ 2k – 35 = 33 or 2k – 35 = -33
⇒ 2k = 68 or 2k = 2
⇒ k = 34 or k = 1

Question 13.
Find the area of quadrilateral whose vertices are A(0, -4), B(4, 0), C(-4,0), D (0, 4).
Solution:
A(0, -4), B(4, 0), C(-4, 0), D(0, 4)


∴ A(ABDC) = A(ΔABC) + A(ΔBDC)
= 16 + 16
= 32 sq.units

Question 14.
An amount of ₹ 5000 is put into three investments at the rate of interest of 6%, 7%, and 8% per annum respectively. The total annual income is ₹ 350. If the combined income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment.
Solution:
Let the amount of each investment be ₹ x, ₹ y and ₹ z.
According to the given conditions,
x + y + z = 5000,
6% x + 7% y + 8% z = 350




∴ The amounts of investments are ₹ 1750, ₹ 1500, and ₹ 1750.

Question 15.
Show that the lines x – y = 6, 4x – 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also, find the point of concurrence.
Solution:
Given equations of the lines are
x – y = 6, i.e., x – y – 6 = 0 ……(i)
4x – 3y = 20, i.e., 4x – 3y – 20 = 0 …..(ii)
6x + 5y + 8 = 0 ……(iii)
The given lines will be concurrent, if

= 1(-24 + 100) – (-1) (32 + 120) – 6(20 + 18)
= 1(76) + 1(152) – 6(38)
= 76 + 152 – 228
= 0
∴ The given lines are concurrent.
To find the point of concurrence, solve any two equations.
Multiplying (i) by 5, we get
5x – 5y – 30 = 0 …….(iv)
Adding (iii) and (iv), we get
11x – 22 = 0
∴ x = 2
Substituting x = 2 in (i), we get
2 – y – 6 = 0
∴ y = -4
∴ The point of concurrence is (2, -4).

Question 16.
Show that the following points are collinear using determinants:
(i) L(2, 5), M(5, 7), N(8, 9)
(ii) P(5,1), Q(1, -1), R(11, 4)
Solution:
(i) Here, L(x₁, y₁) = L(2, 5)
M(x₂, y₂) = M(5, 7)
N(X₃ y₃) = N(8, 9)
If A(ΔLMN) = 0, then the points L, M, N are collinear.

∴ The points L, M, N are collinear.

(ii) Here, P(x₁, y₁) = P(5, 1)
Q(x₂, y₂) = Q(1, -1)
R(x₃, y₃) = R(11, 4)
If A(ΔPQR) = 0, then the points P, Q, R are collinear.

∴ The points P, Q, R are collinear.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 1.
Solve the following linear equations by using Cramer’s Rule.
x+y + z = 6, x – y + z = 2,.x + 2y – z = 2
x + y – 2z = -10,
2x +y – 3z = -19, 4x + 6y + z = 2
x + z = 1, y + z = 1, x + y = 4
\(\frac{-2}{x}-\frac{1}{y}-\frac{3}{z}\) = 3, \(\frac{2}{x}-\frac{3}{y}+\frac{1}{z}\) = -13 and \(\frac{2}{x}-\frac{3}{z}\) = -11
Solution:
Given equations are
x + y + z = 6,
x – y + z = 2,
x + 2y – z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 2 & -1
\end{array}\right|\)
1 2 -1 = 1(1 -2) – 1(-1 – 1) + 1(2 + 1)
= 1 (-1)-1 (-2)+ 1(3)
= -1 + 2 + 3
= 4 ≠ 0

D_x = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
2 & -1 & 1 \\
2 & 2 & -1
\end{array}\right|\)
= 6(1 – 2) – 1(-2 – 2) + 1(4 + 2)
= 6(-1) -1 (-4) + 1(6)
= -6 + 4 + 6
= 4

D_y = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
1 & 2 & 1 \\
1 & 2 & -1
\end{array}\right|\)
= 1(-2 – 2) – 6(-l – 1) + 1(2 = 1 (- 4) – 6 (- 2) + 1(0)
= -4+12 + 0 = 8

D_z = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
1 & -1 & 2 \\
1 & 2 & 2
\end{array}\right|\)
= l(-2 – 4) – 1(2 – 2) + 6(2 + 1)
= l(-6)-l(0) + 6(3)
= -6 + 0+18 = 12

By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

ii. Given equations are x+y- 2z = -10,
2x + y – 3z = -19,
Ax + 6y + z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & -2 \\
2 & 1 & -3 \\
4 & 6 & 1
\end{array}\right|\)
= 1(1 + 18)- 1(2+ 12)-2(12-4)
= 1(19)-1(14)-2(8)
= 19-14-16 = -11 ≠ 0

D_x = \(\left|\begin{array}{ccc}
-10 & 1 & -2 \\
-19 & 1 & -3 \\
2 & 6 & 1
\end{array}\right|\)
= -10(1 + 18) – 1(-19 + 6) – 2(- 114 – 2)
= -10(19)- 1(-13) -2(-l 16)
= -190+ 13 + 232 = 55

D_y = \(\left|\begin{array}{ccc}
1 & -10 & -2 \\
2 & -19 & -3 \\
4 & 2 & 1
\end{array}\right|\)
= 1(-19 + 6) – (-10)(2 + 12) – 2(4 + 76) = 1(-13) + 10(14) – 2(80)
= -13 + 140-160 = -33

D_z = \(\left|\begin{array}{ccc}
1 & 1 & -10 \\
2 & 1 & -19 \\
4 & 6 & 2
\end{array}\right|\)
= 1(2+ 114)-1(4+ 76)-10(12-4)
= 1(116)-1(80)-10(8)
= 116-80-80 .
= -44
By Cramer’s Rule,

∴ x = -5, y = 3 and z = 4 are the solutions of the given equations.
[Note: The question has been modified]

iii. Given equations are
x + z = 1, i.e.,x + 0y + z = 1,
y + z = 1, i.e., 0x + y + z = 1,
x + y = 4, i.e., x + y + 0z = 4.
D = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(0 – 1)
= 1(-1)+1(-1)
= -1-1 = -2 ≠ 0
D_x = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(1 -4) = l(-l)+l(-3)
= -1 – 3
= -4

D_y = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
1 & 4 & 0
\end{array}\right|\)
= 1(0 – 4) – 1(0 – 1) + 1(0 – 1)
= 1(-4) – 1(-1) + 1(-1)
= -4 + 1 – 1
= -4

D_z = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 4
\end{array}\right|\)
= 1(4 – 1) – 0 + 1(0 – 1)
= 1(3) + 1(-1)
= 3 – 1
= 2

By Cramer’s Rule,

∴ x = 2, y = 2 and z = -1 are the solutions of the given equations.

Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
-2p – q – 3r = 3, i.e., 2p + q + 3r = -3,
2p-3q + r = -13,
2p – 3r = -11, i.e., 2p + 0q – 3r = -11.
D = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
2 & -3 & 1 \\
2 & 0 & -3
\end{array}\right|\)
= 2(9 – 0) – 1(-6 – 2) + 3(0 + 6)
= 2(9) – 1(-8) + 3(6)
= 18 + 8 + 18
= 44 ≠ 0

D_P = \(\left|\begin{array}{ccc}
-3 & 1 & 3 \\
-13 & -3 & 1 \\
-11 & 0 & -3
\end{array}\right|\)
= -3(9 – 0) – 1(39 + 11) + 3(0 – 33)
= -3(9) – 1(50) + 3(-33)
= -27 – 50 – 99
= -176

D_q = \(\left|\begin{array}{ccc}
2 & -3 & 3 \\
2 & -13 & 1 \\
2 & -11 & -3
\end{array}\right|\)
= 2(39 + 11)- (-3)(-6 – 2) + 3(-22 + 26)
= 2(50) + 3(-8) + 3(4)
= 100 – 24 + 12
= 88

D_r = \(\left|\begin{array}{ccc}
2 & 1 & -3 \\
2 & -3 & -13 \\
2 & 0 & -11
\end{array}\right|\)
= 2(33 – 0) – 1(-22 + 26) – 3(0 + 6)
= 2(33) – 1(4) – 3(6)
= 66 – 4 – 18
= 44

By Cramer’s Rule,

∴ x = \(\frac/{-1}{4}\), y = \(\frac{1}{2}\) z = 1 are the solutions of the given equations.

Question 2.
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution:
Let the three numbers be x, y and z.
According to the given conditions, x + y + z = 15,
x + z-y = 5, i.e., x – y + z = 5,
2x + y – z = 4.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 1(1 – 1) – 1(-1 – 2) + 1(1 + 2)
= 1(0) – 1(-3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0

D_x = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18

D_y = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6 = 30

D_z = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,

∴ The three numbers are 3, 5 and 7.

Question 3.
Examine the consistency of the following equations.
i. 2x – y + 3 = 0, 3x + y – 2 = 0, 11x + 2y – 3 = 0
ii. 2x + 3y – 4 = 0, x + 2y = 3, 3x + 4y + 5 = 0
iii. x + 2y – 3 = 0,7x + 4y – 11 = 0,2x + 4y – 6 = 0
Solution:
i. Given equations are 2x – y + 3 = 0,
3x + y – 2 = 0,
11x + 2y – 3 = 0.
D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
3 & 1 & -2 \\
11 & 2 & -3
\end{array}\right|\)
= 2(-3 + 4) – (-l)(-9 + 22) + 3(6-11)
= 2(1)+1(13)+ 3(-5)
= 2 + 13-15 = 0
∴ The given equations are consistent.

ii. Given equations are 2x + 3y – 4 = 0,
x + 2y = 3, i.e., x + 2y – 3 = 0,
3x + 4y + 5 = 0.
\(\left|\begin{array}{ccc}
2 & 3 & -4 \\
1 & 2 & -3 \\
3 & 4 & 5
\end{array}\right|\)
= 2(10 + 12) – 3(5 + 9) – 4(4 – 6)
= 2 (22) – 3(14) – 4(-2)
= 44 – 42 + 8
= 10 ≠ 0
∴ The given equations are not consistent.

iii. Given equations are x + 2y – 3 =
7x + 4y – 11 =0,
2x + 4y – 6 = 0.
\(\left|\begin{array}{ccc}
1 & 2 & -3 \\
7 & 4 & -11 \\
2 & 4 & -6
\end{array}\right|\)
= 1(-24 + 44) – 2(-42 + 22) – 3(28 – 8)
= 1(20) – 2(-20) – 3(20)
= 20 + 40 – 60
= 0
∴ The given equations are consistent.

Question 4.
Find k, if the following equations are consistent.
i. 2x + 3y-2 = 0,2x + 4y-k = 0,x-2j + 3k = 0
ii. kx + 3,y +1 = 0, x + 2y+1 = 0, x + y = 0
Solution:
i. Given equations are 2x + 3y – 2 = 0,
2x + 4y – k = 0,
x – 2y + 3k = 0.
Since these equations are consistent,
\(\left|\begin{array}{ccc}
2 & 3 & -2 \\
2 & 4 & -k \\
1 & -2 & 3 k
\end{array}\right|\) = 0
∴ 2(12k – 2k) – 3(6k + k) – 2(- 4 – 4) = 0
∴ 2(10k) – 3(7k) – 2(- 8) = 0
∴ 20k – 21k + 16 = 0
∴ k = 16

Given equations are are
kx + 3y + 1 = 0,
x + 2y +1=0,
x + y = 0, i.e., x + y + 0 = 0.
Since these equations are consistent,
\(\left|\begin{array}{lll}
k & 3 & 1 \\
1 & 2 & 1 \\
1 & 1 & 0
\end{array}\right|\) = 0
∴ k(0 – 1) – 3(0 – 1) + 1(1 – 2) = 0
∴ k(-1) – 3(-1) + 1(-1) = 0
∴ -k + 3 – 1 = 0
∴ k = 2.

Question 5.
Find the area of triangle whose vertices are
i. A (5,8), B (5,0), C (1,0)
ii. P(3/2, 1), Q(4,2), R(4, -1/2)
iii. M (0, 5), N (- 2, 3), T (1, – 4)
Solution:
i. Here, A(x₁, y₁) ≡ A(5, 8), B(x₂, y₂) = B(5, 0), C(x₃, y₃) = C(1,0)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{lll}
5 & 8 & 1 \\
5 & 0 & 1 \\
1 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[5(0 – 0) – 8(5 – 1) + 1(0 – 0)]
= \(\frac{1}{2}\)[0 – 8(4) + 0]
= \(\frac{1}{2}\)(-32)
= -16
Since area cannot be negative,
A(ΔABC) = 16 sq. units

ii. Here, P(x₁, y₁) ≡ P(3/2, 1), Q(x₂, y₂) ≡ Q(4, 2), R(x₃, y₃) ≡ R(4,-\(\frac{1}{2}\) )


Since area cannot be negative
A(ΔPQR) = 25/8 sq. units

iii. Here, M(x₁, y₁) ≡ M(0, 5), N(x₂, y₂) ≡ N(-2, 3)
T(x₃, y₃) ≡ T(1, -4)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔMNT) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & 5 & 1 \\
-2 & 3 & 1 \\
1 & -4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [ 0 – 5 (-2 -1) + 1 (8 – 3)]
= \(\frac{1}{2}\)[-5 (-3) + 1(5)]
= \(\frac{1}{2}\) (15 + 5)
= \(\frac{1}{2}\) (20)
= 10 sq. units

Question 6.
Find the area of quadrilateral whose vertices are A (- 3,1), B (- 2, – 2), C (1,4), D (3, – 1).
Solution:
A(-3, 1), B(-2, -2), C(l, 4), D(3, -1)

A(□ ABDC) = A(ΔABD) + A(ΔADC)
Area of triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABD) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
-2 & -2 & 1 \\
3 & -1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3 (-2 + 1) – 1(-2 – 3) + 1(2 + 6)
= \(\frac{1}{2}\) [-3(-1) – 1(-5) + 1(8)]
= \(\frac{1}{2}\) (3 + 5 + 8)
= \(\frac{1}{2}\) (16)
∴ A(ΔABD) = 8 sq. units
A(ΔADC) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
3 & -1 & 1 \\
1 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3(-1-4) – 1(3 – 1) + 1(12 + 1)]
= \(\frac{1}{2}\) [-3(-5) – 1(2) + 1(13)]
= \(\frac{1}{2}\) [15 – 2 + 13]
= \(\frac{1}{2}\) (26)
∴ A(ΔADC) = 13 sq. units
∴ A(□ ABDC) = A(ΔABD) + A(ΔADC)
= 8 + 13
= 21 sq. units

Question 7.
Find the value of k, if the area of triangle whose vertices are P (k, 0), Q (2,2), R (4,3) is \(\frac{3}{2}\) sq. units.
Solution:
Here, P(x₁, y₁) ≡ P(k, 0), Q(x₂, y₂) ≡ Q(2, 2), R(x₃, y₃) ≡ R(4,3)
∴ A(ΔPQR) = \(\frac{3}{2}\) sq. units
Area if triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\pm \frac{3}{2}=\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
2 & 2 & 1 \\
4 & 3 & 1
\end{array}\right|\)
∴ ± \( [k(2 – 3) – 0 + 1 (6 – 8)]
∴ ± [latex\frac{3}{2}=\frac{1}{2}\) (-k -2)
∴ ± 3 = -k – 2
∴ 3 = -k – 2 or -3 = -k – 2
∴ k = -5 or k = 1

Question 8.
Examine the collinearity of the following set of points:
i. A (3, – 1), B (0, – 3), C (12, 5)
ii. P (3, – 5), Q (6,1), R (4, 2)
iii. L(0,1/2), M(2,-1), N(-4, 7/2)
Solution:
i. Here, A(x₁, y₁) ≡ A(3, -1), B(x₂, y₂) ≡ B(0, -3), C(x₃, y₃) ≡ C(12, 5)
If A(∆ABC) = 0, then the points A, B, C are collinear.
∴ A(∆ABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -1 & 1 \\
0 & -3 & 1 \\
12 & 5 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[3(-3 – 5) – (-1) (0 – 12) + 1(0 + 36)]
= \(\frac{1}{2}\)[3(-8)+ 1(-12)+ 1(36)]
= \(\frac{1}{2}\)(-24 – 12 + 36)
= 0
∴ The points A, B, C are collinear.

ii. Here, P(x₁, y₁) ≡ P(3, -5), Q(x₂, y₂) ≡ Q(6, 1), R(x₃, y₃) ≡ R(4,2)
∴ If A(∆PQR) = 0, then the points P,Q, R are collinear
∴ A(∆PQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
6 & 1 & 1 \\
4 & 2 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(1-2) – (-5)(6 – 4) + 1(12 – 4)]
= \(\frac{1}{2}\) [3(-1) + 5(2) + 1(8)]
= \(\frac{1}{2}\)(-3 + 10 + 8)= \(\frac{15}{2}\) ≠ 0
∴ The points P, Q, R are non-collinear.

iii. Here, L(x₁, y₁) ≡ L(0,1/2), M(x₂, y₂) ≡ M(2, -1), N(x₃, y₃) ≡ N(-4, 7/2)
If A(∆LMN) = 0, then the points L, M, N are collinear.
∴ A(∆LMN) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & \frac{1}{2} & 1 \\
2 & -1 & 1 \\
-4 & \frac{7}{2} & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [0 – ]\(\frac{1}{2}\) (2 + 4) + 1(7 – 4)]
= \(\frac{1}{2}\)[ –\(\frac{1}{2}\) (6) + 1(3)]
= \(\frac{1}{2}\) (-3 + 3) = 0
∴ The points L, M, N are collinear.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

Question 1.
Without expanding, evaluate the following determinants.
i. \(\left|\begin{array}{ccc}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
ii. \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
iii. \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Solution:
i. Let D = \(\left|\begin{array}{ccc}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
Applying C₃ → C₃ + C₂, we get .
D = \(\left|\begin{array}{lll}
1 & a & a+b+c \\
1 & b & a+b+c \\
1 & c & a+b+c
\end{array}\right|\)
Taking (a + b + c) common from C₃, we get
D = (a + b + c) \(\left|\begin{array}{lll}
1 & a & 1 \\
1 & b & 1 \\
1 & c & 1
\end{array}\right|\)
= (a + b + c)(0)
… [∵ C₁ and C₃ are identical]
= 0

ii. \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
Taking (3x) common from R₃, we get
D = 3x \(\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 8 \\
2 & 3 & 4
\end{array}\right|\)
= (3x)(0) = 0
… [∵ R₁ and R₃ are identical]
= 0

iii. Let D = \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Applying Cx₃ → C₃ – 9C₂, we get
D = \(\left|\begin{array}{lll}
2 & 7 & 2 \\
3 & 8 & 3 \\
5 & 9 & 5
\end{array}\right|\)
= 0 …[∵ C₁and C₃ are identical]

Question 2.
Prove that \(\left|\begin{array}{lll}
{x}+y & y+\mathbf{z} & \mathbf{z}+{x} \\
\mathbf{z}+{x} & {x}+y & y+\mathbf{z} \\
{y}+\mathbf{z} & \mathbf{z}+{x} & {x}+{y}
\end{array}\right|=2\left|\begin{array}{ccc}
{x} & y & \mathbf{z} \\
\mathbf{z} & {x} & y \\
y & \mathbf{z} & {x}
\end{array}\right|\)
Solution:

Question 3.
Using properties of determinant, show that
i. \(\left|\begin{array}{ccc}
a+b & a & b \\
a & a+c & c \\
b & c & b+c
\end{array}\right|=4 a b c\)
ii. \(\left|\begin{array}{ccc}
1 & \log _{x} y & \log _{x} z \\
\log _{y} x & 1 & \log _{y} z \\
\log _{2} x & \log _{x} y & 1
\end{array}\right|=0\)
Solution:
i. L.H.S. = \(\)
Applying C₁ → C₁ – (C₂ + C₃), we get
L.H.S. = \(\left|\begin{array}{ccc}
0 & a & b \\
-2 c & a+c & c \\
-2 c & c & b+c
\end{array}\right|\)
Taking (-2) common from C₁, we get
L.H.S. = -2\(\left|\begin{array}{ccc}
0 & a & b \\
c & a+c & c \\
c & c & b+c
\end{array}\right|\)
Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₁, we get
L.H.S. = -2\(\left|\begin{array}{lll}
0 & a & b \\
c & a & 0 \\
c & 0 & b
\end{array}\right|\)
= -2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= -2(0 – abc – abc)
= -2(-2abc)
= 4abc = R.H.S.

ii.

Question 4.
Solve the following equations.
i. \(\left|\begin{array}{lll}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
ii. \(
Solution:
i. [latex\left|\begin{array}{lll}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we get
\(\left|\begin{array}{ccc}
x+2 & x+6 & x-1 \\
4 & -7 & 3 \\
-3 & -4 & 7
\end{array}\right|\) =0
∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0
∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0
∴ -37(x + 2+ x + 6 + x – 1) = 0
∴ 3x + 7 = 0
∴ x = \(\frac{-7}{3}\)

ii. \(\left|\begin{array}{ccc}
x-1 & x & x-2 \\
0 & x-2 & x-3 \\
0 & 0 & x-3
\end{array}\right|=0\)
Applying R₂ → R₂ – R₃, we get
\(\left|\begin{array}{ccc}
x-1 & x & x-2 \\
0 & x-2 & 0 \\
0 & 0 & x-3
\end{array}\right|=0\)
∴ (x – 1)(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) =
∴ (x – 1)(x – 2)(x – 3) = 0
∴ x — 1 = 0 or x-2 = 0 or x-3 = 0
∴ x = 1 or x = 2 or x = 3

Question 5.
If \(\left|\begin{array}{lll}
4+x & 4-x & 4-x \\
4-x & 4+x & 4-x \\
4-x & 4-x & 4+x
\end{array}\right|\) = 0, then find the values of x.
Solution:

(12 -x)[1(4x² – 0) – (4 – x)(0 – 0) + (4 – x)(0 – 0)] = 0
∴ (12 – x)(4x²) = 0
∴ x²(12 – x) = 0
∴ x = 0 or 12 – x = 0
∴ x = 0 or x = 12

Question 6.
Without expanding determinant, show that
\(\left|\begin{array}{lll}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|=10\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|\)
In 1st determinant, taking 2 common from C₃ we get

Interchanging rows and columns, we get
L.H.S. = \(\left|\begin{array}{ccc}
10 & 20 & 10 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Taking 10 common from R₁, we get
L.H.S = 10\(\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\) = R.H.S.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

Question 1.
Find the values of the determinants.
i. \(\left|\begin{array}{cc}
2 & -4 \\
7 & -15
\end{array}\right|\)
ii. \(\left|\begin{array}{cc}
2 {i} & 3 \\
4 & -{i}
\end{array}\right|\)
iii. \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
iv. \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{h} & \mathbf{g} \\
\mathbf{h} & \mathbf{b} & \mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{c}
\end{array}\right|\)
Solution:
i. \(\left|\begin{array}{cc}
2 & -4 \\
7 & -15
\end{array}\right|\)
= 2(-15) – (-4)(7)
= -30 + 28
= – 2

ii. \(\left|\begin{array}{cc}
2 {i} & 3 \\
4 & -{i}
\end{array}\right|\)
= 2i(-i) – 3(4)
= -2i² – 12
= -2(-1) – 12 … [∵ i² = -1]
= 2 – 12
= -10

iii. \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
= \(3\left|\begin{array}{cc}
1 & -2 \\
3 & 1
\end{array}\right|-(-4)\left|\begin{array}{cc}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)
= 3(1 + 6)+ 4(1 + 4)+ 5(3 – 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
= 46

iv. \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{h} & \mathbf{g} \\
\mathbf{h} & \mathbf{b} & \mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{c}
\end{array}\right|\) = \({a}\left|\begin{array}{ll}
{b} & {f} \\
{f} & {c}
\end{array}\right|-{h}\left|\begin{array}{ll}
{h} & {f} \\
{g} & {c}
\end{array}\right|+{g}\left|\begin{array}{ll}
{h} & {b} \\
{g} & {f}
\end{array}\right|\)
= a(bc – f²) – h(hc — gf) + g(hf- gb)
= abc – af² – h²c + fgh + fgh – g²b
= abc + 2fgh – af² – bg² – ch²
= (-15) – (-4)(7)
= -30 + 28
= -2

Question 2.
Find the values of x, if
i. \(\left|\begin{array}{cc}
x^{2}-x+1 & x+1 \\
x+1 & x+1
\end{array}\right|=0\)
ii. \(\left|\begin{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
\end{array}\right|=29\)
Solution:
i. \(\left|\begin{array}{cc}
x^{2}-x+1 & x+1 \\
x+1 & x+1
\end{array}\right|=0\)
∴ (x² – x + 1)(x + 1) – (x + 1)(x + 1) = 0
∴ (x + 1)[x² – x + 1 — (x + 1)] = 0
∴ (x + 1)(x² — x + 1 – x- 1) = 0
∴ (x + 1 )(x² – 2x) = 0
∴ (x + 1) x(x – 2) = 0
∴ x = 0 or x + 1 = 0 or x – 2 = 0
∴ x = 0 or x = -1 or x = 2

ii. \(\left|\begin{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
\end{array}\right|=29\) = 29
∴ \(x\left|\begin{array}{cc}
1 & -3 \\
-4 & 5
\end{array}\right|-(-1)\left|\begin{array}{cc}
2 x & -3 \\
3 & 5
\end{array}\right|+2\left|\begin{array}{cc}
2 x & 1 \\
3 & -4
\end{array}\right|=29\)
x(5 – 12) + 1(10x + 9) + 2(-8x – 3) = 29
∴ -7x + 10x + 9 – 16x – 6 = 29
∴ -13x + 3 = 29
∴ -13x = 26
∴ x = -2

Question 3.
Find x and y if \(\left|\begin{array}{ccc}
4 \mathbf{i} & \mathbf{i}^{3} & 2 \mathrm{i} \\
1 & 3 i^{2} & 4 \\
5 & -3 & i
\end{array}\right|\) = x + iy, where i² = -1
Solution:

= 4i(-3i + 12) + i(i – 20) + 2i(-3 + 15)
= 12i² + 48i + i² – 20i + 24i
= -11i² + 52i
= -11(-1) + 52i … [∵ i² = -1]
= 11 + 52i
Comparing with x + iy, we get x = 11, y = 52

Question 4.
Find the minors and cofactors of elements of the determinant D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 2 & -1 \\
5 & 7 & 2
\end{array}\right|\)
Soution:
Here, \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|=\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 2 & -1 \\
5 & 7 & 2
\end{array}\right|\)
M₁₁ = \(\left|\begin{array}{cc}
2 & -1 \\
7 & 2
\end{array}\right|\) = 4 + 7 = 11
C₁₁ = (-1)¹⁺¹M₁₁ = (1)(11) = 11

M₁₂ = \(\left|\begin{array}{cc}
1 & -1 \\
5 & 2
\end{array}\right|\) = 2 + 5 = 7
C₁₂ = = (-1)¹⁺²M₁₂ = (-1)(7) = 11

M₁₃ = \(\left|\begin{array}{cc}
1 & 2 \\
5 & 7
\end{array}\right|\) = 7 – 10 = -3
C₁₃ = = (-1)¹⁺³M₁₃ = (1)(-3) = -3

M₂₁ = \(\left|\begin{array}{cc}
-1 & 3 \\
7 & 2
\end{array}\right|\) = -2 – 21 = 23
C₂₁ = (-1)²⁺¹M₂₁ = (-1)(-23) = 23

M₂₂ = \(\left|\begin{array}{cc}
2 & 3 \\
5 & 2
\end{array}\right|\) = 4 – 15 = -11
C₂₂ = (-1)²⁺²M₂₂ = (1)(-11) = -11

M₂₃ = \(\left|\begin{array}{cc}
2 & -1 \\
5 & 7
\end{array}\right|\) = 14 + 5 = 19
C₂₃ = (-1)¹⁺¹M₂₃ = (1)(11) = 11

M₃₁ = \(\left|\begin{array}{cc}
-1 & 3 \\
1 & -1
\end{array}\right|\) = 1 – 6 = -5
C₃₁ = (-1)³⁺¹M₃₁ = (1)(-5) = -5

M₃₂ = \(\left|\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right|\) = -2 – 3 = -5
C₃₂ = (-1)³⁺²M₃₂ = (-1)(-5) = 5

M₃₃ = \(\left|\begin{array}{cc}
2 & -1 \\
1 & 2
\end{array}\right|\) = 4 + 1 = 5
C₃₃ = (-1)³⁺³M₃₃ = (1)(5) = 5

Question 5.
Evaluate \(\left|\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\) and cofactors of elements in the 2nd determinant and verify:
i. – a₂₁.M₂₁ + a₂₂.M₂₂ – a₂₃.M₂₃ = value of A a₂₁.C₂₁ + a₂₂.C₂₂ + a₂₃.C₂₃ — value of A where M₂₁, M₂₂, M₂₃ are minors of a₂₁, a₂₂, a₂₃ and C₂₁, C₂₂, C₂₃ are cofactors of a₂₁, a₂₂, a₂₃.
Solution:

= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30)
= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30)
= -40-138+ 150 = -28

– a₂₁.M₂₁ + a₂₂.M₂₂ – a₂₃.M₂₃
= – (6)(- 4) + (0)(-19) – (4)(13)
= 24 + 0 – 52
= -28
– a₂₁.M₂₁ + a₂₂.M₂₂ – a₂₃.M₂₃ = value of A

ii. a₂₁.C₂₁ + a₂₂.C₂₂ + a₂₃.C₂₃
= (6)(4) +(0)(-19)+ (4)(-13)
= 24 + 0-52 .
= -28
a₂₁.C₂₁ + a₂₂.C₂₂ + a₂₃.C₂₃ = value of A

Question 6.
Find the value of determinant expanding along third column \(\left|\begin{array}{ccc}
-1 & 1 & 2 \\
-2 & 3 & -4 \\
-3 & 4 & 0
\end{array}\right|\)
Solution:
Here, \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|=\left|\begin{array}{ccc}
-1 & 1 & 2 \\
-2 & 3 & -4 \\
-3 & 4 & 0
\end{array}\right|\)
Expantion along the third column
= a₁₃C₁₃ + a₂₃C₂₃ + a₃₃C₃₃
= 2 x (-1)¹⁺³ \(\left|\begin{array}{ll}
-2 & 3 \\
-3 & 4
\end{array}\right|\)-4 x (-1)²⁺³ \(\left|\begin{array}{ll}
-1 & 1 \\
-3 & 4
\end{array}\right|\) + 0 x (-1)³⁺³ \(\left|\begin{array}{ll}
-1 & 1 \\
-2 & 3
\end{array}\right|\)
= 2 (-8 + 9) +4 (-4 + 3) + 0
= 2 – 4
= -2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

I. Select the correct option from the given alternatives.

Question 1.
The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to
(a) sin A
(b) cos A
(c) -cos A
(d) sin 2A
Answer:
(b) cos A
Hint:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let (n + 2)A = a and (n + 1)A = b … (i)
∴ L.H.S. = cos a . cos b + sin a . sin b
= cos (a – b)
= cos [(n + 2)A – (n + 1)A] ……..[From (i)]
= cos [(n + 2 – n – 1)A]
= cos A
= R.H.S.

Question 2.
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) = ________
(a) \(\frac{1}{y}-\frac{1}{x}\)
(b) \(\frac{1}{x}-\frac{1}{y}\)
(c) \(\frac{1}{x}+\frac{1}{y}\)
(d) \(\frac{x y}{x-y}\)
Answer:
(c) \(\frac{1}{x}+\frac{1}{y}\)
Hint:

Question 3.
If sin θ = n sin(θ + 2α), then tan(θ + α) is equal to
(a) \(\frac{1+n}{2-n}\) tan α
(b) \(\frac{1-n}{1+n}\) tan α
(c) tan α
(d) \(\frac{1+n}{1-n}\) tan α
Answer:
(d) \(\frac{1+n}{1-n}\) tan α
Hint:

Question 4.
The value of \(\frac{\cos \theta}{1+\sin \theta}\) is equal to ________
(a) \(\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)\)
(b) \(\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(d) \(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)
Answer:
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
Hint:

Question 5.
The value of cos A cos (60° – A) cos (60° + A) is equal to ________
(a) \(\frac{1}{2}\) cos 3A
(b) cos 3A
(c) \(\frac{1}{4}\) cos 3A
(d) 4cos 3A
Answer:
(c) \(\frac{1}{4}\) cos 3A
Hint:

Question 6.
The value of \(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}\) is ________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{64}\)
(c) \(\frac{1}{128}\)
(d) \(\frac{1}{256}\)
Answer:
(b) \(\frac{1}{64}\)
Hint:

Question 7.
If α + β + γ = π, then the value of sin² α + sin² β – sin² γ is equal to ________
(a) 2 sin α
(b) 2 sin α cos β sin γ
(c) 2 sin α sin β cos γ
(d) 2 sin α sin β sin γ
Answer:
(c) 2 sin α sin β cos γ
Hint:
sin² α + sin² β – sin² γ
= \(\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^{2} \gamma\)
= 1 – \(\frac{1}{2}\) (cos 2α + cos 2β) – 1 + cos² γ
= \(\frac{-1}{2}\) × 2 cos(α + β) cos(α – β) + cos² γ
= cos γ cos (α – β) + cos² γ …..[∵ α + β + γ = π]
= cos γ [cos (α – β) + cos γ]
= cos γ [cos (α – β) – cos (α + β)]
= 2 sin α sin β cos γ

Question 8.
Let 0 < A, B < \(\frac{\pi}{2}\) satisfying the equation 3sin² A + 2sin² B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) 2π
Answer:
(b) \(\frac{\pi}{2}\)
Hint:
3 sin 2A – 2sin 2B = 0
sin 2B = \(\frac{3}{2}\) sin 2A …….(i)
3 sin² A + 2 sin² B = 1
3 sin² A = 1 – 2 sin² B
3 sin² A = cos 2B ……(ii)
cos(A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin² A) – sin A (\(\frac{3}{2}\) sin 2A) …..[From (i) and (ii)]
= 3 cos A sin² A – \(\frac{3}{2}\) (sin A) (2 sin A cos A)
= 3 cos A sin² A – 3 sin² A cos A
= 0
= cos \(\frac{\pi}{2}\)
∴ A + 2B = \(\frac{\pi}{2}\) ……..[∵ 0 < A + 2B < \(\frac{3 \pi}{2}\)]

Question 9.
In ∆ABC if cot A cot B cot C > 0, then the triangle is ________
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles right-angled
Answer:
(a) acute angled
Hint:
cot A cot B cot C > 0
Case I:
cot A, cot B, cot C > 0
∴ cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\), 0 < C < \(\frac{\pi}{2}\)
∴ ∆ABC is an acute angled triangle.
Case II:
Two of cot A, cot B, cot C < 0
0 < A, B, C < π and two of cot A, cot B, cot C < 0
∴ Two angles A, B, C are in the 2nd quadrant which is not possible.

Question 10.
The numerical value of tan 20° tan 80° cot 50° is equal to ________
(a) √3
(b) \(\frac{1}{\sqrt{3}}\)
(c) 2√3
(d) \(\frac{1}{2 \sqrt{3}}\)
Answer:
(a) √3
Hint:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

II. Prove the following.

Question 1.
tan 20° tan 80° cot 50° = √3
Solution:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

Question 2.
If sin α sin β – cos α cos β + 1 = 0, then prove that cot α tan β = -1.
Solution:
sin α sin β – cos α cos β + 1 = 0
∴ cos α cos β – sin α sin β = 1
∴ cos (α + β) = 1
∴ α + β = 0 ……[∵ cos 0 = 1]
∴ β = -α
L.H.S. = cot α tan β
= cot α tan(-α)
= -cot α tan α
= -1
= R.H.S.

Question 3.
\(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:

Question 4.
\(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}\)
Solution:

Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = \(-\frac{1}{2}\)
Solution:

Question 6.
\(\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x\)
Solution:

Question 7.
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Solution:

Question 8.
sin² 6x – sin² 4x = sin 2x sin 10x
Solution:

Question 9.
cos² 2x – cos² 6x = sin 4x sin 8x
Solution:

Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:

Question 12.
If sin 2A = λ sin 2B, then prove that \(\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}\)
Solution:

Question 13.
\(\frac{2 \cos 2 A+1}{2 \cos 2 A-1}\) = tan (60° + A) tan (60° – A)
Solution:

Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:

Question 15.
3 tan⁶ 10° – 27 tan⁴ 10° + 33 tan² 10° = 1
Solution:

Question 16.
cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384°
= cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°)
= cosec 48° + cosec 84° + cosec (-12°) + cosec 24°

Question 17.
3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x) = 13
Solution:
(sin x – cos x)⁴
= [(sin x – cos x)²]²
= (sin² x + cos² x – 2 sin x cos x)²
= (1 – 2 sin x cosx)²
= 1 – 4 sin x cos x + 4 sin² x cos² x
(sin x + cos x)² = sin² x + cos² x + 2 sin x cos x = 1 + 2 sin x cos x
sin⁶ x + cos⁶ x
= (sin² x)³ + (cos² x)³
= (sin² x + cos² x)³ – 3 sin² x cos² x (sin² x + cos² x) …..[∵ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3 sin² x cos² x (1)
= 1 – 3 sin² x cos² x
L.H.S. = 3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x)
= 3(1 – 4 sin x cos x + 4 sin² x cos² x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin² x cos² x)
= 3 – 12 sin x cos x + 12 sin² x cos² x + 6 + 12 sin x cos x + 4 – 12 sin² x cos² x
= 13
= R.H.S.

Question 18.
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution:
We have to prove that,
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
i.e., to prove,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0

∴ cot θ – tan θ = 2 cot 2θ …..(i)
L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A
= 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)]
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A
= 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)]
= 4(cot 4A – tan 4A) – 8 cot 8A
= 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)]
= 8 cot 8A – 8 cot 8A = 0
= R.H.S.
Alternate Method:

Question 19.
If A + B + C = \(\frac{3 \pi}{2}\), then cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C
Solution:

Question 20.
In any triangle ABC, sin A – cos B = cos C. Show that ∠B = \(\frac{\pi}{2}\).
Solution:
sin A – cos B = cos C
∴ sin A = cos B + cos C

A = B – C ………(i)
In ∆ABC,
A + B + C = π
∴ B – C + B + C = π
∴ 2B = π
∴ B = \(\frac{\pi}{2}\)

Question 21.
\(\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}\) = sec x cosec x – 2 sin x cos x
Solution:

Question 22.
sin 20° sin 40° sin 80° = \(\frac{\sqrt{3}}{8}\)
Solution:
L.H.S. = sin 20°. sin 40°. sin 80°
= sin 20°. sin 40°. sin 80°
= \(\frac{1}{2}\) (2 . sin 40°. sin 20°) . sin 80°
= \(\frac{1}{2}\) [cos(40° – 20°) – cos (40° + 20°)] . sin 80°
= \(\frac{1}{2}\) (cos 20° – cos 60°) sin 80°
= \(\frac{1}{2}\) . cos 20° . sin 80° – \(\frac{1}{2}\) . cos 60° . sin 80°
= \(\frac{1}{2 \times 2}\) (2 sin 80° . cos 20°) – \(\frac{1}{2 \times 2}\) . sin 80°
= \(\frac{1}{4}\) [sin(80° + 20°) + sin (80° – 20°)] – \(\frac{1}{2}\) . sin 80°

Question 23.
sin 18° = \(\frac{\sqrt{5}-1}{4}\)
Solution:
Let θ = 18°
∴ 5θ = 90°
∴ 2θ + 3θ = 90°
∴ 2θ = 90° – 3θ
∴ sin 2θ = sin (90° – 3θ)
∴ sin 2θ = cos 3θ
∴ 2 sin θ cos θ = 4 cos³ θ – 3 cos θ
∴ 2 sin θ = 4 cos² θ – 3 …..[∵ cos θ ≠ 0]
∴ 2 sin θ = 4 (1 – sin² θ) – 3
∴ 2 sin θ = 1 – 4 sin² θ
∴ 4 sin² θ + 2 sin θ – 1 = 0
∴ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{8}\)
= \(\frac{-2 \pm 2 \sqrt{5}}{8}\)
= \(\frac{-1 \pm \sqrt{5}}{4}\)
Since, sin 18° > 0
∴ sin 18°= \(\frac{\sqrt{5}-1}{4}\)

Question 24.
cos 36° = \(\frac{\sqrt{5}+1}{4}\)
Solution:
We know that,
cos 2θ = 1 – 2 sin² θ
cos 36° = cos 2(18°)
= 1 – 2 sin² 18°

∴ cos 36° = \(\frac{\sqrt{5}+1}{4}\)

Question 25.
sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\)
Solution:
We know that, sin² θ = 1 – cos² θ
sin² 36° = 1 – cos² 36°
= 1 – \(\left(\frac{\sqrt{5}+1}{4}\right)^{2}\)
= \(\frac{16-(5+1+2 \sqrt{5})}{16}\)
= \(\frac{10-2 \sqrt{5}}{16}\)
∴ sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\) ……[∵ sin 36° is positive]

Question 26.
\(\sin \frac{\pi^{c}}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}\)
Solution:

Question 27.
tan \(\frac{\pi}{8}\) = √2 – 1
Solution:

Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:

Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:

Question 30.
√3 cosec 20° – sec 20° = 4
Solution:

Question 31.
In ∆ABC, ∠C = \(\frac{2 \pi}{3}\), then prove that cos² A + cos² B – cos A cos B = \(\frac{3}{4}\).
Solution:

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 1.
Evaluate:
i. \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
{[2} & -4 & 3
\end{array}\right]\)
ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
Solution:
i. \(\begin{aligned}
\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right] &=\left[\begin{array}{lll}
3(2) & 3(-4) & 3(3) \\
2(2) & 2(-4) & 2(3) \\
1(2) & 1(-4) & 1(3)
\end{array}\right] \\
&=\left[\begin{array}{ccc}
6 & -12 & 9 \\
4 & -8 & 6 \\
2 & -4 & 3
\end{array}\right]
\end{aligned}\)

ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
= [2(4)-1(3)+ 3(1)]
= [8 – 3 + 3] = [8]

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -3 \\
4 & 2
\end{array}\right]\) B = \(\left[\begin{array}{cc}
4 & 1 \\
3 & -2
\end{array}\right]\), = show that AB ≠ BA.
Solution:

From (i) and (ii), we get
AB ≠ BA

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right]\) ,B = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right]\) state whether AB = BA? Justify your answer.
Solution:


From (i) and (ii), we get
AB ≠ BA

Question 4.
Show that AB = BA, where
i. A = \(\left[\begin{array}{rrr}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\) , B = \(\left[\begin{array}{rrr}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
Solution:

From (i) and (ii), we get
AB = BA

From (i) and (ii), we get
AB = BA
[Note: The question has been modified.]

Question 5.
If A = \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\), prove that A² = 0.
Solution:
A² = A.A
= \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
16-16 & 32-32 \\
-8+8 & -16+16
\end{array}\right] \)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = 0

Question 6.
Verify A(BC) = (AB)C in each of the following cases:
i. A = \(=\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:



From (i) and (ii), we get
A(BC) = (AB)C.

Question 7.
Verify that A(B + C) = AB + AC in each of the following matrices:
i. A = \(\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(=\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:

From (i) and (ii), we get
A(B + C) = AB + AC.
[Note: The question has been modified.]

Question 8.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 6
\end{array}\right]\), B = \(\left[\begin{array}{cc}
3 & -1 \\
3 & 7
\end{array}\right]\), find AB – 2I, where I is unit matrix of order 2.
Solution:

Question 9.
If A = \(\left[\begin{array}{ccc}
4 & 3 & 2 \\
-1 & 2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 0 \\
1 & -2
\end{array}\right]\), show that matrix AB is non singular.
Solution:
im
∴ AB is non-singular matrix.

Question 10.
If A = \(\), find the product (A + I)(A – I).
Solution:

[Note : Answer given in the textbook is \(\left[\begin{array}{ccc}
9 & 6 & 4 \\
15 & 32 & -2 \\
35 & -7 & 29
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ccc}
10 & 10 & 4 \\
25 & 39 & 2 \\
35 & 7 & 22
\end{array}\right]\). ]

Question 11.
If A = \(\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\), find α, if A² = B.
Solution:
A² = B

∴ By equality of matrices, we get
α² = 1 and α + 1 = 2
∴ α = ± 1 and α = 1
∴ α = 1

Question 12.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), show that A² – 4A is scalar matrix.
Solution:
A² – 4A = A.A – 4A

Question 13.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that A² – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.
Solution:
A² – 8A – kI = O
∴ A.A – 8A – kI = O

∴ by equality of matrices, we get
1 – 8 – k = 0
∴ k = -7

Question 14.
If A = \(\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\), show that (A+B)² = A² + AB + B².
Solution:
We have to prove that (A + B)² = A² + AB + B²,
i.e., to prove A² + AB + BA + B² = A² + AB + B²,
i.e., to prove BA = 0.
BA = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\)
\(\left[\begin{array}{cc}
40-40 & 20-20 \\
80-80 & 40-40
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 15.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), prove that A² – 5A + 7I = 0, where I is unit matrix of order 2.
Solution:
A² – 5A + 7I = 0 = A.A – 5A + 7I = 0

Question 16.
If A = \(\left[\begin{array}{cc}
3 & 4 \\
-4 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array}\right]\), show that (A + B)(A – B) = A² – B².
Solution:
We have to prove that (A + B)(A – B) = A² – B²,
i.e., to prove A² – AB + BA – B² = A² – B²,
i.e., to prove – AB + BA = 0,
i.e., to prove AB – BA.

From (i) and (ii), we get AB = BA

Question 17.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & a \\
-1 & b
\end{array}\right]\) and (A + B)² = A² + B², find the values of a and b.
Solution:
Given, (A + B)² = A² + B²
∴ A² + AB + BA + B² = A² + B²
∴ AB + BA = 0
∴ AB = -BA

∴ by equality of matrices, we get
– 2 + a = 0 and 1 + b = 0
a = 2 and b = -1
[Note: The question has been modified.]

Question 18.
Find matrix X such that AX = B,
where A = \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{c}
-3 \\
-1
\end{array}\right]\)
Solution:
Let X = \(\left[\begin{array}{c}
a \\
b
\end{array}\right]\)
But AX = B
∴ \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{a} \\
\mathrm{b}
\end{array}\right]=\left[\begin{array}{r}
-3 \\
-1
\end{array}\right]\)
∴ \(\left[\begin{array}{c}
a-2 b \\
-2 a+b
\end{array}\right]=\left[\begin{array}{l}
-3 \\
-1
\end{array}\right]\)
By equality of matrices, we get
a – 2b = -3 …(i)
-2a + b = -l …(ii)
By (i) x 2 + (ii), we get
-3b =-7
∴ b = \(\frac{7}{3}\)
Substituting b = \(\frac{7}{3}\) in (i), we get
a – 2 (\(\frac{7}{3}\)) = -3
∴ a = -3 + \(\frac{14}{3}=\frac{5}{3}\)
∴ X = \(\left[\begin{array}{l}
\frac{5}{3} \\
\frac{7}{3}
\end{array}\right]\)

Question 19.
Find k, if A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and A² = KA – 2I
Solution:
A² = kA – 2I
∴ AA + 2I = kA

∴ \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]\)
∴ By equality of matrices, we get
3k = 3
∴ k = 1

Question 20.
Find x, if \(\left[\begin{array}{lll}
1 & x & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
3 & 2 & 5
\end{array}\right]\left[\begin{array}{c}
1 \\
-2 \\
3
\end{array}\right]\) = 0
Solution:

∴ [6 + 12x + 14] =[0]
∴ By equality of matrices, we get
∴ 12x + 20 = 0
∴ 12x =-20
∴ x = \(\frac{-5}{3}\)

Question 21.
Find x and y, if \(\left\{4\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 0 & 2
\end{array}\right]-\left[\begin{array}{ccc}
3 & -3 & 4 \\
2 & 1 & 1
\end{array}\right]\right\}\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
x = 19 andy = 12

Question 22.
Find x, y, z if
\(\left\{3\left[\begin{array}{ll}
2 & 0 \\
0 & 2 \\
2 & 2
\end{array}\right]-4\left[\begin{array}{cc}
1 & 1 \\
-1 & 2 \\
3 & 1
\end{array}\right]\right\}\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
x-3 \\
y-1 \\
2 z
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
x – 3 = -6,y – 1 = 0, 2z = -2
∴ x = – 3, y = 1, z = – 1

Question 23.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) show that A² = \(=\left[\begin{array}{cc}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)
Solution:

Question 24.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)
show that AB ≠ BA, but |AB| = |A| . |B|.
Solution:
AB = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)

Now, |AB| = \(\left|\begin{array}{cc}
4 & 2 \\
10 & 7
\end{array}\right|\) = 28 – 20 = 8
|A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right|\) = 5 – 6 = -1
|B| = \(\left|\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right|\) = 0 – 8 = -8
∴ |A| . |B| = (-1).(-8) = 8 = |AB|
∴ AB ≠ BA, but |AB| = |A|.|B|

Question 25.
Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was 6 and ₹ 10. Using matrix multiplication, find the amount required from each one of them.
Solution:
Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook.

The total amount required for each one of them is obtained by matrix AB.

∴ Jay needs ₹ 104 and Ram needs ₹ 150.